Further maths vector question

Watch
Announcements
Thread starter 7 months ago
#1
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
0
7 months ago
#2
(Original post by alulaaustralis)
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
Note that they aren't giving a vector equation for the line AB, they are giving the particular vector that takes you from A to B. The t in this expression is not a variable t that can take any value, it is one particular value that gives you the vector from A to B.

What you say about AB being perpendicular to the lines is of course true, and this is what they do in the next stage (taking the scalar product of the vector from A to B with the direction vector of the lines) to work out the particular value of t that makes this work.
0
Thread starter 7 months ago
#3
(Original post by Pangol)
Note that they aren't giving a vector equation for the line AB, they are giving the particular vector that takes you from A to B. The t in this expression is not a variable t that can take any value, it is one particular value that gives you the vector from A to B.

What you say about AB being perpendicular to the lines is of course true, and this is what they do in the next stage (taking the scalar product of the vector from A to B with the direction vector of the lines) to work out the particular value of t that makes this work.
I see, so they have simply taken A away from B hence the t meaning μ - λ ?
Thank you so much!
0
7 months ago
#4
(Original post by alulaaustralis)
I don't understand the first line of working. Why does the vector equation of AB have the same direction vector (5,4,3) as the two lines, when AB should technically be perpendicular to those lines and have a different direction vector? What am I missing here?
In the first line you set up a vector for the distance AB by determining the position and direction vector. So far, so good. For a direction vector you need a parameter in form of t. The general parameter t can be got by the difference of the Greek letters given in the first line (can't see them in attachment) or read out in the equation for r. To get the concrete value of parameter t, you calculate the scalar product for perpendicular direction. The rest is a piece of cake: put the value t in your vector AB, get the coordinates and as last step you calculate the length of the vector with these coordinates.
0
7 months ago
#5
(Original post by alulaaustralis)
I see, so they have simply taken A away from B hence the t meaning μ - λ ?
Thank you so much!
Yeah, that's exactly it. I can see why it looks confusing, as λ and μ are used as variable parameters in the definition of the lines themselves, but at the start of their working they mean to take the particular values of λ and μ that get you to A and B. So when they then say that t = μ - λ, that is one particular value of t that gives you the vector from A to B, which they then go on to determine.
0
X

new posts
Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Poll

Join the discussion

Are you tempted to change your firm university choice on A-level results day?

Yes, I'll try and go to a uni higher up the league tables (31)
29.25%
Yes, there is a uni that I prefer and I'll fit in better (10)
9.43%
No I am happy with my choice (59)
55.66%
I'm using Clearing when I have my exam results (6)
5.66%